YES Termination proof of AG_#3.24_rand.trs

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRStoRelADPProof (⇔, 0 ms)
↳2 RelADPP
↳3 RelADPDepGraphProof (⇔, 0 ms)
↳4 RelADPP
↳5 RelADPCleverAfsProof (⇒, 53 ms)
↳6 QDP
↳7 MRRProof (⇔, 0 ms)
↳8 QDP
↳9 QDPOrderProof (⇔, 10 ms)
↳10 QDP
↳11 QDPOrderProof (⇔, 0 ms)
↳12 QDP
↳13 PisEmptyProof (⇔, 0 ms)
↳14 YES

(0) Obligation:

Relative term rewrite system:
The relative TRS consists of the following R rules:

f(0) → s(0)
f(s(0)) → s(0)
f(s(s(x))) → f(f(s(x)))

The relative TRS consists of the following S rules:

rand(x) → rand(s(x))
rand(x) → x

(1) RelTRStoRelADPProof (EQUIVALENT transformation)

We upgrade the RelTRS problem to an equivalent Relative ADP Problem [IJCAR24].

(2) Obligation:

Relative ADP Problem with
absolute ADPs:

f(0) → s(0)
f(s(0)) → s(0)
f(s(s(x))) → F(f(s(x)))
f(s(s(x))) → f(F(s(x)))

and relative ADPs:

rand(x) → RAND(s(x))
rand(x) → x

(3) RelADPDepGraphProof (EQUIVALENT transformation)

We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
1 SCC with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 1 subproblem.

(4) Obligation:

Relative ADP Problem with
absolute ADPs:

f(0) → s(0)
f(s(s(x))) → f(F(s(x)))
f(s(s(x))) → F(f(s(x)))
f(s(0)) → s(0)

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(5) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:s_1 = f_1 = 0 = F_1 = rand_1 =
Found this filtering by looking at the following order that orders at least one DP strictly:Recursive path order with status [RPO].
Quasi-Precedence:

0 > [F1, s1, f1]

Status:
F1: multiset
s1: multiset
f1: multiset
0: multiset

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F0(s0(s0(x))) → F0(f0(s0(x)))
F0(s0(s0(x))) → F0(s0(x))

The TRS R consists of the following rules:

f0(00) → s0(00)
f0(s0(s0(x))) → f0(f0(s0(x)))
rand0(x) → rand0(s0(x))
f0(s0(00)) → s0(00)
rand0(x) → x

Q is empty.
We have to consider all (P,Q,R)-chains.

(7) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(F0(x1)) = x1   
POL(f0(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F0(s0(s0(x))) → F0(f0(s0(x)))
F0(s0(s0(x))) → F0(s0(x))

The TRS R consists of the following rules:

f0(00) → s0(00)
f0(s0(s0(x))) → f0(f0(s0(x)))
rand0(x) → rand0(s0(x))
f0(s0(00)) → s0(00)

Q is empty.
We have to consider all (P,Q,R)-chains.

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


F0(s0(s0(x))) → F0(s0(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F0(x1)  =  F0(x1)
s0(x1)  =  s0(x1)
f0(x1)  =  f0(x1)
00  =  00
rand0(x1)  =  rand0

Recursive path order with status [RPO].
Quasi-Precedence:
00 > [F01, s01, f01, rand0]

Status:
F01: multiset
s01: multiset
f01: multiset
00: multiset
rand0: multiset


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

f0(00) → s0(00)
f0(s0(s0(x))) → f0(f0(s0(x)))
rand0(x) → rand0(s0(x))
f0(s0(00)) → s0(00)

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F0(s0(s0(x))) → F0(f0(s0(x)))

The TRS R consists of the following rules:

f0(00) → s0(00)
f0(s0(s0(x))) → f0(f0(s0(x)))
rand0(x) → rand0(s0(x))
f0(s0(00)) → s0(00)

Q is empty.
We have to consider all (P,Q,R)-chains.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


F0(s0(s0(x))) → F0(f0(s0(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F0(x1)  =  x1
s0(x1)  =  s0(x1)
f0(x1)  =  f0
00  =  00
rand0(x1)  =  rand0

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s0_1=3
00=1
f0=5
rand0=2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

f0(00) → s0(00)
f0(s0(s0(x))) → f0(f0(s0(x)))
rand0(x) → rand0(s0(x))
f0(s0(00)) → s0(00)

(12) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f0(00) → s0(00)
f0(s0(s0(x))) → f0(f0(s0(x)))
rand0(x) → rand0(s0(x))
f0(s0(00)) → s0(00)

Q is empty.
We have to consider all (P,Q,R)-chains.

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) YES